K vs Kp  [ENDORSED]

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Anushi Patel 1J
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K vs Kp

Postby Anushi Patel 1J » Thu Jan 10, 2019 6:52 pm

If an equation involves gases only and we are given the equilibrium constant K, can we assume that they mean Kp?

Matthew Tran 1H
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Joined: Fri Sep 28, 2018 12:16 am

Re: K vs Kp

Postby Matthew Tran 1H » Thu Jan 10, 2019 7:20 pm

I believe that since the book typically uses partial pressures for homogeneous reactions with gases, the p in Kp is implied when they give K. It would be safe to assume that if they give you a value of K and give measures of partial pressure for homogeneous reactions with gases that K=Kp.

Lisa Werner 2F
Posts: 72
Joined: Fri Sep 28, 2018 12:17 am

Re: K vs Kp  [ENDORSED]

Postby Lisa Werner 2F » Thu Jan 10, 2019 8:52 pm

Yes, you can assume Kp as essentially the only way to find the K of homogenous gas is through the partial pressure. Similarly, you can assume Kc when dealing with the K of homogenous aqueous reactions because you use molarity to find it.

Parth Mungra
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Joined: Fri Sep 28, 2018 12:28 am

Re: K vs Kp

Postby Parth Mungra » Thu Jan 10, 2019 9:51 pm

What's the difference between the two? They are calculated the same way, but Kp doesn't include reactants/products in the aqueous/solid state?

Vicky Lu 1L
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Re: K vs Kp

Postby Vicky Lu 1L » Thu Jan 10, 2019 10:18 pm

K in general is an expression to describe the composition of the reaction at equilibrium. Kp is for concentration of partial pressure of the reactants/products while Kc is used in concentration of the reactants/products. Solid and liquids aren't included in the equilibrium constant expression, only gases or aqueous solutions. Kc is used in molar concentration while Kp is for partial pressure.


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